Answer

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**Hint:**It is given that angles A and B are complementary angles. We know that the sum of two complementary angles is ${{90}^{\circ }}$. So, we can write summation of A and B as ${{90}^{\circ }}$. After that we can write either A or B in terms of another angle. And then we will substitute angle B in terms of A in the above options and see which option is correct.

**Complete step by step answer:**

The two angles given in the above problem are A and B and these angles are complementary angles.

We know that the sum of any two complementary angles is ${{90}^{\circ }}$ so we can write the sum of angles A and B as ${{90}^{\circ }}$.

Writing the sum of angles A and B as ${{90}^{\circ }}$ we get,

$\angle A+\angle B={{90}^{\circ }}$

Writing angle B in terms of angle A in the above equation we get,

$\angle B={{90}^{\circ }}-\angle A$

Substituting the above relation in all the options given in the above equation so first of all we are going to substitute in option (a) we get,

$\begin{align}

& \sin A=\sin B \\

& \Rightarrow \sin A=\sin \left( {{90}^{\circ }}-A \right) \\

\end{align}$

We know that $\sin \left( {{90}^{\circ }}-\theta \right)=\cos \theta $ so using this relation in the above equation we get,

$\sin A=\cos A$

In the above equation, $L.H.S\ne R.H.S$ so option (a) is not correct.

Checking option (b) we get,

$\begin{align}

& \cos A=\cos B \\

& \Rightarrow \cos A=\cos \left( {{90}^{\circ }}-A \right) \\

\end{align}$

We know that $\cos \left( {{90}^{\circ }}-\theta \right)=\sin \theta $ so using this relation in the above we get,

$\cos A=\sin A$

Again, $L.H.S\ne R.H.S$ so option (b) is not correct.

Checking option (c) we get,

$\begin{align}

& \tan A=\tan B \\

& \Rightarrow \tan A=\tan \left( {{90}^{\circ }}-A \right) \\

\end{align}$

We know that $\tan \left( {{90}^{\circ }}-\theta \right)=\cot \theta $ so using this relation in the above we get,

$\tan A=\cot A$

Again, $L.H.S\ne R.H.S$ so option (c) is not correct.

Now, checking the option (d) we get,

$\begin{align}

& \sec A=\csc B \\

& \Rightarrow \sec A=\csc \left( {{90}^{\circ }}-A \right) \\

\end{align}$

We know that $\csc \left( {{90}^{\circ }}-\theta \right)=\sec \theta $ so using this relation in the above we get,

$\sec A=\sec A$

Again, $L.H.S=R.H.S$ so option (d) is correct.

**So, the correct answer is “Option d”.**

**Note:**To solve the above problem, you must have knowledge of the complementary angle conversion of trigonometric ratios otherwise we cannot move forward in this problem. Also, you should know what complementary angles means. This is also required in this problem.

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